| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 This is a standard C3 trigonometric identities question with three progressive parts: (a) routine proof using given identity, (b) standard half-angle identity derivation, (c) equation solving requiring substitution and algebraic manipulation. While it requires multiple techniques, all are well-practiced C3 content with no novel insights needed. Slightly above average due to the multi-step nature and need to connect parts (a) and (b) to solve (c). |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05l Double angle formulae: and compound angle formulae1.05m Geometric proofs: of trig sum and double angle formulae1.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| let \(A = B = \frac{x}{2}\) \(\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}\) | M1 | |
| \(\cos x = (1-\sin^2\frac{x}{2}) - \sin^2\frac{x}{2}\) | M1 | |
| \(\cos x = 1 - 2\sin^2\frac{x}{2}\) | A1 | |
| (b) LHS \(= \frac{1-(1-2\sin^2\frac{x}{2})}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \tan\frac{x}{2} =\) RHS | M1 A1 | |
| (c) \(\tan\frac{x}{2} = 2\sec^2\frac{x}{2} - 5\), \(\tan\frac{x}{2} = 2(1+\tan^2\frac{x}{2}) - 5\) | M1 | |
| \(2\tan^2\frac{x}{2} - \tan\frac{x}{2} - 3 = 0\), \((2\tan\frac{x}{2} - 3)(\tan\frac{x}{2} + 1) = 0\) | M1 | |
| \(\tan\frac{x}{2} = -1\) or \(\frac{3}{2}\) | A1 | |
| \(\frac{x}{2} = 135\) or \(56.310\) | B1 | |
| \(x = 112.6°\) (1dp), \(270°\) | A2 | (12 marks) |
**(a)** $\cos(A+B) = \cos A \cos B - \sin A \sin B$
let $A = B = \frac{x}{2}$ $\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$ | M1 |
$\cos x = (1-\sin^2\frac{x}{2}) - \sin^2\frac{x}{2}$ | M1 |
$\cos x = 1 - 2\sin^2\frac{x}{2}$ | A1 |
**(b)** LHS $= \frac{1-(1-2\sin^2\frac{x}{2})}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \tan\frac{x}{2} =$ RHS | M1 A1 |
**(c)** $\tan\frac{x}{2} = 2\sec^2\frac{x}{2} - 5$, $\tan\frac{x}{2} = 2(1+\tan^2\frac{x}{2}) - 5$ | M1 |
$2\tan^2\frac{x}{2} - \tan\frac{x}{2} - 3 = 0$, $(2\tan\frac{x}{2} - 3)(\tan\frac{x}{2} + 1) = 0$ | M1 |
$\tan\frac{x}{2} = -1$ or $\frac{3}{2}$ | A1 |
$\frac{x}{2} = 135$ or $56.310$ | B1 |
$x = 112.6°$ (1dp), $270°$ | A2 | (12 marks)\begin{enumerate}[label=(\alph*)]
\item Use the identity
$$\cos (A + B) = \cos A \cos B - \sin A \sin B$$
to prove that
$$\cos x \equiv 1 - 2 \sin^2 \frac{x}{2}.$$ [3]
\item Prove that, for $\sin x \neq 0$,
$$\frac{1 - \cos x}{\sin x} \equiv \tan \frac{x}{2}.$$ [3]
\item Find the values of $x$ in the interval $0 \leq x \leq 360^{\circ}$ for which
$$\frac{1 - \cos x}{\sin x} = 2 \sec^2 \frac{x}{2} - 5,$$
giving your answers to 1 decimal place where appropriate. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q7 [12]}}